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General Tabletop Discussion
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probabilty part Deux
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<blockquote data-quote="The Sigil" data-source="post: 345635" data-attributes="member: 2013"><p>In this case, you have an exercise in permutations.</p><p></p><p>Let us look at how many permuations there are (a permutation is where rolling 1 then 2 then 3 then 4 is different from 1 then 2 then 4 then 3 - if they're considered the same, that's a combination).</p><p></p><p>You have 20^4 permuations (each die can come up in 20 different ways and there are four dice), or 160,000 permutations.</p><p></p><p>How many of those permutations contain at least two 1's? Easy.</p><p></p><p>1-1-X-X has 400 possible permutations (the two X dice can come up in 20 different ways - including more 1's - and there are two of them, so it's 20^2).</p><p></p><p>1-X-1-X has another 400 (800 total)</p><p>1-X-X-1 has another 400 (1200 total)</p><p>X-1-1-X has another 400 (1600 total)</p><p>X-1-X-1 has another 400 (2000 total)</p><p>X-X-1-1 has another 400 (2400 total).</p><p></p><p>Note that we have already included combinations with 3 and/or 4 ones in the above total. You would have X coming up in 19 different ways (instead of 20) if you wanted EXACTLY 2 rolls of 1 (instead of AT LEAST 2 rolls of 1) since in that case you would want only rolls of 2 through 20. In that case you would have instead of 6*(20^2) the total of 6*(19^2) or 2166 total possible permutations.</p><p></p><p>Our final fraction for AT LEAST two 1's is then 2400/160,000 or 0.015 (corresponding to 1.5%).</p><p></p><p>Our final fraction for EXACTLY two 1's is 2166/160,00 or 0.0135375 (which is about 1.35%).</p><p></p><p>--The Sigil</p></blockquote><p></p>
[QUOTE="The Sigil, post: 345635, member: 2013"] In this case, you have an exercise in permutations. Let us look at how many permuations there are (a permutation is where rolling 1 then 2 then 3 then 4 is different from 1 then 2 then 4 then 3 - if they're considered the same, that's a combination). You have 20^4 permuations (each die can come up in 20 different ways and there are four dice), or 160,000 permutations. How many of those permutations contain at least two 1's? Easy. 1-1-X-X has 400 possible permutations (the two X dice can come up in 20 different ways - including more 1's - and there are two of them, so it's 20^2). 1-X-1-X has another 400 (800 total) 1-X-X-1 has another 400 (1200 total) X-1-1-X has another 400 (1600 total) X-1-X-1 has another 400 (2000 total) X-X-1-1 has another 400 (2400 total). Note that we have already included combinations with 3 and/or 4 ones in the above total. You would have X coming up in 19 different ways (instead of 20) if you wanted EXACTLY 2 rolls of 1 (instead of AT LEAST 2 rolls of 1) since in that case you would want only rolls of 2 through 20. In that case you would have instead of 6*(20^2) the total of 6*(19^2) or 2166 total possible permutations. Our final fraction for AT LEAST two 1's is then 2400/160,000 or 0.015 (corresponding to 1.5%). Our final fraction for EXACTLY two 1's is 2166/160,00 or 0.0135375 (which is about 1.35%). --The Sigil [/QUOTE]
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